Question

Chapter 2, Problem 2/153 (video solution to similar problem attached) At the bottom of a loop in the vertical (r, θ) plane at an altitude of 390 m, the airplane P has a horizontal velocity of 700 km/h and no horizontal acceleration. The radius of curvature of the loop is 1255 m. For the radar tracking at o, determine the recorded values of r and θ for this instant. 700 km/h 390 m 1095 m Answers: m/s2 rad/s2

Answer 1

Now, r = squareroot 390^2 + 1095^2 = 1162 m now the angle is theta = tan^-1(390/1095) = 19. 60 degrees given v = 700 times 1095/4900 = 156. 43 now a = a_n = v_2/row = 156. 43/1255 = 19. 50 m/sec^2 v_r = r = v cos theta = 156.43 cos 19. 60 = 147. 37 m/s v_e = r theta - 156. 43 sin 19. 60 = 1162 theta theta = -0. 0452 rad/sec now a-r = r - r theta^2 = a sin theta = 19.50 sin 19. 60 + 1162 (-0. 0452)^2 = 8. 915 m/s^2 \r\n a_e = r theta + 2 r theta = a cos theta = 19. 50 cos 19. 60 - 2(147. 37) - 0. 0452/1162 = 0. 0282 rad/s^2